3.18 \(\int \frac{(a+b x^2)^2 (A+B x^2)}{x^5} \, dx\)

Optimal. Leaf size=51 \[ -\frac{a^2 A}{4 x^4}-\frac{a (a B+2 A b)}{2 x^2}+b \log (x) (2 a B+A b)+\frac{1}{2} b^2 B x^2 \]

[Out]

-(a^2*A)/(4*x^4) - (a*(2*A*b + a*B))/(2*x^2) + (b^2*B*x^2)/2 + b*(A*b + 2*a*B)*Log[x]

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Rubi [A]  time = 0.0383888, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {446, 76} \[ -\frac{a^2 A}{4 x^4}-\frac{a (a B+2 A b)}{2 x^2}+b \log (x) (2 a B+A b)+\frac{1}{2} b^2 B x^2 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(A + B*x^2))/x^5,x]

[Out]

-(a^2*A)/(4*x^4) - (a*(2*A*b + a*B))/(2*x^2) + (b^2*B*x^2)/2 + b*(A*b + 2*a*B)*Log[x]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2 (A+B x)}{x^3} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (b^2 B+\frac{a^2 A}{x^3}+\frac{a (2 A b+a B)}{x^2}+\frac{b (A b+2 a B)}{x}\right ) \, dx,x,x^2\right )\\ &=-\frac{a^2 A}{4 x^4}-\frac{a (2 A b+a B)}{2 x^2}+\frac{1}{2} b^2 B x^2+b (A b+2 a B) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0244401, size = 50, normalized size = 0.98 \[ b \log (x) (2 a B+A b)-\frac{a^2 \left (A+2 B x^2\right )+4 a A b x^2-2 b^2 B x^6}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(A + B*x^2))/x^5,x]

[Out]

-(4*a*A*b*x^2 - 2*b^2*B*x^6 + a^2*(A + 2*B*x^2))/(4*x^4) + b*(A*b + 2*a*B)*Log[x]

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Maple [A]  time = 0.006, size = 51, normalized size = 1. \begin{align*}{\frac{{b}^{2}B{x}^{2}}{2}}+A\ln \left ( x \right ){b}^{2}+2\,B\ln \left ( x \right ) ab-{\frac{A{a}^{2}}{4\,{x}^{4}}}-{\frac{abA}{{x}^{2}}}-{\frac{{a}^{2}B}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(B*x^2+A)/x^5,x)

[Out]

1/2*b^2*B*x^2+A*ln(x)*b^2+2*B*ln(x)*a*b-1/4*a^2*A/x^4-a/x^2*A*b-1/2*a^2/x^2*B

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Maxima [A]  time = 0.964059, size = 73, normalized size = 1.43 \begin{align*} \frac{1}{2} \, B b^{2} x^{2} + \frac{1}{2} \,{\left (2 \, B a b + A b^{2}\right )} \log \left (x^{2}\right ) - \frac{A a^{2} + 2 \,{\left (B a^{2} + 2 \, A a b\right )} x^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^5,x, algorithm="maxima")

[Out]

1/2*B*b^2*x^2 + 1/2*(2*B*a*b + A*b^2)*log(x^2) - 1/4*(A*a^2 + 2*(B*a^2 + 2*A*a*b)*x^2)/x^4

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Fricas [A]  time = 1.45611, size = 122, normalized size = 2.39 \begin{align*} \frac{2 \, B b^{2} x^{6} + 4 \,{\left (2 \, B a b + A b^{2}\right )} x^{4} \log \left (x\right ) - A a^{2} - 2 \,{\left (B a^{2} + 2 \, A a b\right )} x^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^5,x, algorithm="fricas")

[Out]

1/4*(2*B*b^2*x^6 + 4*(2*B*a*b + A*b^2)*x^4*log(x) - A*a^2 - 2*(B*a^2 + 2*A*a*b)*x^2)/x^4

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Sympy [A]  time = 0.644465, size = 49, normalized size = 0.96 \begin{align*} \frac{B b^{2} x^{2}}{2} + b \left (A b + 2 B a\right ) \log{\left (x \right )} - \frac{A a^{2} + x^{2} \left (4 A a b + 2 B a^{2}\right )}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(B*x**2+A)/x**5,x)

[Out]

B*b**2*x**2/2 + b*(A*b + 2*B*a)*log(x) - (A*a**2 + x**2*(4*A*a*b + 2*B*a**2))/(4*x**4)

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Giac [A]  time = 1.10515, size = 97, normalized size = 1.9 \begin{align*} \frac{1}{2} \, B b^{2} x^{2} + \frac{1}{2} \,{\left (2 \, B a b + A b^{2}\right )} \log \left (x^{2}\right ) - \frac{6 \, B a b x^{4} + 3 \, A b^{2} x^{4} + 2 \, B a^{2} x^{2} + 4 \, A a b x^{2} + A a^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^5,x, algorithm="giac")

[Out]

1/2*B*b^2*x^2 + 1/2*(2*B*a*b + A*b^2)*log(x^2) - 1/4*(6*B*a*b*x^4 + 3*A*b^2*x^4 + 2*B*a^2*x^2 + 4*A*a*b*x^2 +
A*a^2)/x^4